find-lim-x-x-2-e-1-x-e-1-x-1-

Question Number 30504 by abdo imad last updated on 22/Feb/18

f i n d {l i m}_{x \to \infty} x^{2} (e^{\frac{1}{x}} - e^{\frac{1}{x + 1}}) .

Answered by sma3l2996 last updated on 24/Feb/18

{\underset{x \to \infty}{l i m x}}^{2} (e^{1 / x} - e^{1 / (x + 1)})

e^{1 / x} \underset{\infty}{\sim} 1 + \frac{1}{x} + \frac{1}{2 x^{2}} a n d e^{1 / (x + 1)} \underset{\infty}{\sim} 1 + \frac{1}{x + 1} + \frac{1}{2 {(x + 1)}^{2}}

s o e^{1 / x} - e^{1 / (x + 1)} \underset{\infty}{\sim} \frac{1}{x} + \frac{1}{2 x^{2}} - \frac{1}{x + 1} - \frac{1}{2 (x^{2} + 1)} = \frac{1}{x (x + 1)} - \frac{1}{2 x^{2} (x^{2} + 1)}

{\underset{x \to \infty}{l i m x}}^{2} (e^{1 / x} + e^{1 / (x + 1)}) = \underset{x \to \infty}{l i m} (\frac{x^{2}}{x (x + 1)} - \frac{x^{2}}{2 x^{2} (x^{2} + 1)})

= \underset{x \to \infty}{l i m} \frac{x}{x + 1} = 1